Relationship between wild ramification and restriction in Galois extension For each $n \geq 1$, we denote by $\mathbb{Q}_n$ the unique subfield of $\mathbb{Q}(\zeta_{p^{n+1}})$ for which $[\mathbb{Q}_n:\mathbb{Q}]=p^n$. Let now $K$ be a finite Galois extension of $\mathbb{Q}$ with the following properties:


*

*K contains $\mathbb{Q}_m$ for some $m \geq 1$;

*There exists an element $g \in Gal(K/\mathbb{Q})$ with the property that the order of $g$ is $p^n$ with $n>m$ and $\left. g \right|_{\mathbb{Q}_m}$ is the generator of the Galois group $Gal(\mathbb{Q}_m/\mathbb{Q})$. 


Is it then true that $K$ must contain $\mathbb{Q}_t$ for some $t>m$? 
I have tried to use quite a lot of theory around wild ramification and local considerations (including this post wildly ramified extensions and $p$-power roots of unity ), without success in producing a proof/counterexample.
 A: Since the prime $p$ ($\neq 2$ for simplification) plays a special role, let us concentrate on the case when $K/Q$ is a $p$-extension, i.e. a Galois extension whose group $G$ is a finite $p$-group. The case of a general $G$ can be derived from this provided we have sufficient information on the structure of $G$. Suppose also that $K/Q$ is $S$-ramified, i.e. unramified outside a finite set of primes $S$={$p$} $\cup$ $T$, with $T$ = {$l_1,...,l_r$} or $T$ empty. Note that the presence of $p$ is necessary because we'll deal with the so-called cyclotomic $Z_p$-extension of $Q$, namely $G_S (k) = \cup Q_n$. 
For a number field $k$ and the set $S_k$ of places of $k$ above $S$ (we'll write $S$ for $S_k$, no confusion is possible), denote by $k_S$ the maximal pro-$p$-extension of $k$ which is $S$-ramified, and $G_S (k) = Gal(k_S/k)$. The description of $G_S (k)$ by generators and relations can be approached using CFT and Galois cohomology, as for example in H. Koch's book "Galois Theory of p-extensions". In particular, the minimal number $d$ of generators (resp. the number $r$ of relations) of $G_S (k)$ is equal to the $Z/pZ$-dimension of $H^1(G_S (k), Z/pZ)$ (resp. of $H^2(.)$). Actually, $d$ has been computed by Shafarevitch (op. cit.,§11, thm. 11.8), and $r$ is given by a so-called Euler-Poincaré characteristic formula : $1-d+r = -c$, where $c$ is the number of complex places of $k$.
After these preliminaries, we can attack your problem over $k = Q$. Since no ramification condition is specified, we must consider two  separate cases:
1) If $T$ is empty, i.e. only wild ramification is allowed, Shafarevitch's formula over $Q$ gives $d=1$ and $r=0$, in other words, $G_S (Q)$ is the free pro-$p$-group on a single generator, which is $Z_p$. Equivalently, $Q_S = Q_{\infty}$ , which implies that  $K$ = a certain  $Q_n$ , and the answer to your question is trivially YES
2) If $T$ is not empty, i.e. tame ramification is allowed, let $ Q_m = K \cap Q_{\infty}$ , $G = Gal(K/Q)$ and $H = Gal(K/Q_m)$. Suppose, as you do, that there exists $g \in G$, of order $n>m$, such that its image $ g$ mod $H$ in $G/H$ generates $G/H = Gal(Q_m/Q)$. Replace $K$ by the fixed field $L$ of $(H \cap<g>)$, so that the image $g$ mod $(H\cap<g>)$ in $Gal(L/Q)$ generates a subgroup isomorphic to $Gal(Q_m/Q)$. This means that that the exact sequence $1 \to Gal(L/Q_m) \to Gal(L/Q) \to Gal(Q_m/Q) \to 1$ splits, i.e. describes a semi-direct product. So in this case the answer to your question will be NO if we can construct the adequate extension $L/Q_m$ . If $T$ = {$l_1,...,l_r$}, the number of relations of $G_S (Q)$ will be $r$, more precisely the relations will come from $r$ local tame relations of the type (op. cit., §10, thm. 10.2) $t^{l-1}[t^{-1}, s^{-1}]=1$ (necessarily, $l\equiv 1$ mod $p$ to allow ramification). This means that the maximal abelian quotient of $G_S (Q)$, considered as a $Z_p$-module, will have rank 1 and torsion isomorphic to a product of cyclic groups $Z/(l -1)Z$. By adjusting our $l$, we can immediately construct the desired extension $L/Q_m$  ./.
A: On re-reading my second comment, I do not find it very enlightening, so I propose to replace it by this more detailed answer (mainly in order to be able to type math symbols (*)). I also edited two unfortunate misprints in my first answer (just before the exact sequence of Galois groups).
So let us construct our abelian extension $K/Q$ with the required $g$ of order $p^n$, with $n>m$, and such that $Gal(K/Q)\cong Gal(K/Q_m) \times Gal(Q_m/Q)$. We have seen that he Galois group of the maximal abelian $S$-ramified $S$-extension of $Q$ is a direct product ($Z_p, +$)$\times \mathcal T$, where the   torsion group $\mathcal T$ is a finite product of cyclic groups of the form $Z/(l-1)Z$. Because a subgroup of $Z_p$ is isomorphic to $Z_p$ hence is pro-$p$-free, $Gal(K/Q)$ will be viewed as a quotient of $Gal(Q_m/Q) \times\mathcal T$ and $Gal(K/Q_m)$ as a quotient of $\mathcal T$. Then choose  $l \in T$, such that $p^n$ divides exactly $ 1- p^n$, so that $Gal(K/Q)$ will contain an element $g$ of order $p^n$ coming from $\mathcal T$. Taking further $l' \in T, l' \neq l $,  $T=${$l, l'$}, and $Gal(K/Q_m) \cong$ the factor $Z/(l'-1)Z$ of $\mathcal T$, we get our extension $K$ with the desired properties. Things could be seen much more clearly by drawing Galois diagrams.
